Can anyone answer part 3 for me? I've tried so many incorrect answers that I've already given up: 1.00, 1.50, 2.00, 3.00, 9.40
Question: A slingshot obeying Hooke's law is used to launch pebbles vertically into the air. You observe that if you pull a pebble back 2.00 cm against the elastic band, the pebble goes 3.00 m high.
Part A (already answered correctly): Assuming that air drag is negligible, how high will the pebble go if you pull it back 5.00 cm instead? Answer: h = 18.8 m
Part B (already answered correctly): How far must you pull it back so it will reach 13.0 m? Answer: x = 4.16 cm
Part C: If you pull a pebble that is twice as heavy back 3.00 cm, how high will it go? Answer: h = ___ cm
given 3m high, at 2cm pull.
OK, you are pulling it 50 percent more, and doubling mass.
doubling mass will halve the height (PE=mgh)
The PE in the elastic goes by 1/2 k x^2, so if you increase the x to 1.50, that squared is 2.25x, so
doubling mass, pulling to 3cm, should make height 3m*1/2*2.25
To solve part C, we can use the concept of conservation of energy.
First, let's calculate the potential energy of the pebble at its highest point when pulled back 2.00 cm. We can use the formula for potential energy, which is given by:
Potential Energy = mass * acceleration due to gravity * height
Since the pebble goes 3.00 m high, we have:
Potential Energy1 = m * g * h1
Potential Energy1 = m * 9.8 m/s^2 * 3.00 m
Now, let's consider the pebble that is twice as heavy. The potential energy at its highest point when pulled back 3.00 cm can be calculated using the same formula:
Potential Energy2 = (2m) * 9.8 m/s^2 * h2
Since we want to find the height, we can set the potential energies equal to each other:
Potential Energy1 = Potential Energy2
m * 9.8 m/s^2 * 3.00 m = (2m) * 9.8 m/s^2 * h2
Simplifying the equation:
3.00 m = 2 * h2
Divide both sides by 2:
h2 = 3.00 m / 2
h2 = 1.50 m
Therefore, the pebble will go 1.50 m high if you pull it back 3.00 cm.
To answer Part C of the question, you need to apply the concepts and equations related to Hooke's law and the conservation of energy. Let's break it down step by step:
1. Start by understanding the key principles involved. Hooke's law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. In this case, pulling the pebble back against the elastic band represents a displacement.
2. Recall that the potential energy stored in a spring can be calculated using the equation: PE = 0.5 * k * x^2, where PE is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.
3. Since the mass of the pebble has doubled, the force required to pull it back will also double, but the spring constant remains the same. Therefore, the potential energy stored in the spring will be twice as much as in the original scenario.
4. Use the conservation of energy principle to equate the potential energy stored in the spring to the gravitational potential energy at the maximum height of the pebble. This can be expressed as: PE = m * g * h, where m is the mass of the pebble, g is the acceleration due to gravity, and h is the maximum height reached.
5. Rearrange the equation to solve for h: h = PE / (m * g).
6. Substitute the relevant values into the equation. The potential energy (PE) is doubled in this scenario, while the mass (m) and acceleration due to gravity (g) remain constant.
7. Calculate the height (h) using the equation and the given values.
By following these steps, you can find the answer to Part C of the question. Remember to double-check your calculations and units to ensure accuracy.